Poj3187backward digit sums (DFS)

Link: HuangJing
Train of Thought: It's easy to think of DFS, but it's so hard to calculate those values like Yang Hui's triangle. I think my teammates are really amazing, recursively calculate the value at the top of the Yang Hui triangle .... For detailed implementation, see the code...
Question:

Language:DefaultBackward digit sums Time limit:1000 ms   Memory limit:65536 K Total submissions:4285   Accepted:2474 Description FJ and his cows enjoy playing a mental game. they write down the numbers from 1 to n (1 <= n <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. they repeat this until only a single number is left. for example, one instance of the game (when n = 4) might go like this: 3 1 2 4 4 3 6 7 9 16 Behind FJ's back, the cows have started playing a more difficult game, in which they try to determine the Starting sequence from only the final total and the number n. unfortunately, the game is a bit above FJ's mental arithmetic capabilities. Write a program to help FJ play the game and keep up with the cows. Input Line 1: two space-separated integers: N and the final sum. Output Line 1: An ordering of the integers 1 .. n that leads to the given sum. if there are multiple solutions, choose the one that is lexicographically least, I. E ., that puts smaller numbers first. Sample Input 4 16 Sample output 3 1 2 4 Hint Explanation of the sample: There are other possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest. Source Usaco 2006 February Gold & Silver

Code:

#include<cstdio>#include<cstring>const int maxn=10+10;int a[maxn],ans[maxn],n,target,flag;bool vis[maxn];int forever(int n){    if(n==0)  return a[0];    for(int i=0;i<n-1;i++)        a[i]=a[i]+a[i+1];    return forever(n-1);}void dfs(int pos){    if(flag)  return;    if(pos==n)    {        for(int i=0;i<=n-1;i++)            a[i]=ans[i];        if(forever(n)==target)        {            flag=1;            for(int i=0;i<n-1;i++)                printf("%d ",ans[i]);            printf("%d\n",a[n-1]);        }    }    for(int i=1;i<=n;i++)    {        if(!vis[i])        {            vis[i]=true;            ans[pos]=i;            dfs(pos+1);            vis[i]=false;        }    }}int main(){    while(~scanf("%d%d",&n,&target))    {       flag=0;       memset(vis,false,sizeof(vis));       for(int i=1;i<=n;i++)       {           ans[0]=i;           vis[i]=true;           dfs(1);           vis[i]=false;       }    }    return 0;}